# Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14D Solutions

## Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14D Solutions

### Exercise 14D

(Q1) Find the slope and y – intercept of the line:

(i) y = 4

(ii) ax – by = 0

(iii) 3x – 4y = 5

Solution:

(i) Y = 4

General equation, y = mx + c

Given equation is y = 4

Y = 0x + 4

m = slope

∴ m = 0

And y – intercept = 4

(ii) ax – by = 0

General equation, y = mx + c

We have to convert given equation in general equation.

ax – by = 0

ax = by

a/b x = y

y = a/b x + 0

∴ m = a/b and y intercept = 0

(iii) 3x – 4y = 5

General equation, y = mx + c

We have to convert given equation in general equation.

3x – 4y = 5

3x = 5 + 4y

3x – 5 = 4y

3x – 5/4 = y

Y = 3x/4 -5/4

∴ Slope (m) = ¾ and

y – Intercept = -5/4

(Q2) The equation of a line x – y = 4. Find its slope and y-intercept also find its inclination.

Solution:

Given equation of line is x – y = 4

We have to find slope (m) =? Y-Intercept =? Inclination (θ) = ?

General equation, y = mx + c

We have to convert given equation in general form,

x – y = 4

x – 4 = y

y = x – 4

∴ Slope (m) = 1 and

y – Intercept = -4

We know that,

m = tan θ

∴ tan θ  = 1

tan θ = tan 45°

∴ θ = 45°

∴ Inclination (θ) = 45°

(Q3) (i)  Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0?

(ii) Is the line x – 3y = 4

Perpendicular to the line 3x – y =?

(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y =?

(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.

Solution:

If the two lines are perpendicular then the slope is-

Given lines are 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0

Now, we have to find slope of these lines –

First line => 3x + 4y + 7= 0

4y = -3x -7

Y = -3x -7/4

Y = (-3/4) x + (-7/4)

Comparing this equation with the general form of y = mx + c

∴ m1 = -3/4

Second line: 28x – 21y + 50 = 0

-21y = -28x – 50

y = -28x-50/-21

y = -28/-21 x + (-50/-20)

y = 28/21 x + (50/21)

y = (4/3) x + (50/21)

comparing this equation with general equation y = mx + c

∴ m2 = 4/3

We know that,

m1 × m2 = -1

(-3/4) × (4/3) = -1

-12/12 = -1

1 = -1

∴ Two lines are perpendicular

(ii) If the two lines are perpendicular then the slope is ‘-1’ that is,

i.e. m1 × m2 = 1

Given lines are x – 3y = 4 and 3x – y  = 7

Now, we have to find slope of these lines-

First line => x – 3y = 4

-3y = 4 – x

Y = 4-x/-3

Y = 4/-3 + (-x/-3)

= x/3 + (4/-3)

Y = 1/3 x + (4/-3)

Comparing this equation with the general form of y = mx + c

∴ m1 =1/3

Second line => 3x – y = 7

-y = 7 – 3x

y = -7 + 3x

y = 3x – 7

y = 3x + (-7)

Comparing this equation with general form of y = mx + c

∴ m2 = 3

∴ We know that,

m1 × m2 = -1

3×1/3 ≠ -1

∴ the slopes of two lines are 1 not equal to -1 .

Hence the two lines are not perpendicular to each other.

(iii) If the two lines are parallel then the slope are same.

Given lines are 3x + 2y = 5 and x + 2y = 1

Now, we have to find slope of these lines-

First line=> 3x + 2y = 5

2y = 5-3x

2y = (-3)x + 5

Y = (-3)x + 5/2

Y = (-3/2)x + 5/2

Comparing this equation with the general form of y = mx + c

∴ m1 = -3/2

∴ Second line => x + 2y = 1

2y = 1-x

2y = -x +1

y = -x+1/2

y = (-1/2)x + ½

Comparing this equation with the general form of y = mx + c

∴ m2 = -1/2

We can see that the slope of m1 and m2 are not equal or same.

Hence, the two lines are not parallel to each other

(iv) Given that, slope (m) = 2

We know that,

y2 – y1 /x2 – x1 = m

(1, 4) = (x1, y1) and (x, 2) = (x2, y2)

2-4/x-1 = 2

-2 = 2× (x -1)

-2 = 2x -2

-2+2 = 2x

0 = 2x

x = 0

(Q4) Find the slope of the line which is parallel to:

(i) x + 2y + 3 = 0

(ii) x/2 – y/3 – 1= 0

Solution:

Given equation of line is x+2y + 3 = 0

First we have to convert given equation in general form y = mx + c.

X + 2y + 3 = 0

2y = -x-3

y = -x-3/2

y = -x/2 + (-3/2)

y = -1/2 x + (-3/2)

Comparing this equation with y = mx + c

∴ m = -1/2

(ii) Given equation of line is x/2 – y/3 – 1 = 0

First we have to convert given equation with y = mx + c

x/3 – y/3 -1 = 0

-y/3 = -x/3 +1

Multiply by ‘3’ on both sides –

3 × (-y/3) = 3 × (-x/3) + 3

-y = -3x + 3

y = 3x – 3

Comparing this equation with y = mx + c

∴ Slope (m) = 3

(Q5) Find the slope of the line which is perpendicular to:

(i) x – y/2 + 3 = 0

(ii) x/3 – 2y = 4

Solution:

Given equation of line is

x – y/2 + 3= 0

First we have to convert given equation in general form of y = mx + c

x –y/2 + 3 = 0

-y/2 = -3 – x

-y = 2(-3 –x)

-y = – 6 – 2x

y = 6 + 2x

y = 2x + 6

Comparing with y = mx +c

∴ Slope (m1) = 2

The slope of lines is perpendicular is –

m1 × m2 =- -1

m2 = -1/m1

m2 = -1/2

Slope (m2) = -1/2

(ii) Given equation of line is

x/3 – 2y = 4

First we have to convert given equation in general form of y = mx + c

x/3 – 2y = 4

-2y = 4 – x/3

y = -2 (4- x/3)

= -8 + 2x/3

y = 2/3 x + (-8)

Comparing this equation with y = mx + c

∴ Slope (m1) = 2/3

The slope of line is perpendicular to the line-

m1 × m2 = -1

m2 = -1/m1

m2 = -1/(2/3)

m2 = -3/2

(Q6) Find the value of P if the lines, whose equations are 2x – y + 5 and px + 3y = 4 are perpendicular to each other.

Solution:

Given equations are –

2x – y + 5 = 0 —– (1) and

Px + 3y = 4 —– (2)

First we have to convert equation (1) into general form of y = mx + c

2x – y + 5 = 0

-y = – 2x – 5

y = 2x +5

Comparing this equation with y = mx + c

∴ Slope (m1) = 2

Now, we have to convert equation (2) in general form of y = mx + c

Px + 3y = 4

3y = 4 – px

3y = – px + 4

y = – px + 4/3

y = -p/3 x + 4/3

Comparing this equation with y = mx + c

∴ Slope (m2) = -p /3

If two lines are perpendicular then the product of slopes of two line is -1

m1 × m2 = -1

2 × -p/3 = -1

-2p/3 = -1

2p/3 = 1

P = 3/2

(Q7)  The equation of a line AB is 2x – 2y + 3 = 0

(i) Find the slope of the line AB

(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.

Solution:

Given equation of line is 2x – 2y + 3 = 0

(i) First we have to find slope of this line.

So, now we have to convert given equation of line into the general form of y = mx + c

2x – 2y + 3 = 0

-2y = -2x -3

2y = 2x +3

y = 2x + 3/2

y = 2x/2 + 3/2

y = x + 3/2

Comparing this equation with y = mx + c

Slope (m) = 1

∴ Slope of line AB = 1

(ii) The slope of line AB is 1

So, we know that the value of tan 45° = 1

∴ We know that,

Slope = tan θ

= 1

And tan 45° = 1

∴ θ = 45°

(Q8) The lines represented by 4x +3y = 9 and px – 6y + 3 = 0 are parallel find the value of p.

Solution:

Given equation of lines are 4x + 3y = 9 —– (1)

And px – 6y + 3 = 0

First we have to convert equation (1) in general form of y = mx + c

4x + 3y = 9

3y = 9 – 4x

3y = – 4x + 9

y = – 4x+9/3

y = -4x/3 + 9/3

y = -4/3 x + 9/3

∴ Slope (m1) = -4/3

Now, we have to convert equation (2) in general form of y = mx + c

Px – 6y + 3 = 0

-6y = -3 – px

-6y = – px -3

y = px-3/ -6

y = -px/-6 + (-3/-6)

= px/6 + 3/6

Y = px/6 + ½

∴ Slope (m2) = p/6

Also given that two lines are parallel.

We know that, if two line are parallel then the slopes of two lines are same or equal.

∴ m1 = m2

-4/3 = p/6

-4 = 3p/6

-4 = p/2

-8 = p

P = -8

(Q9) If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.

Solution:

Given lines are y = 3x + 7 —— (1)

And 2y + px = 3 —— (2)

First we have to find slope

y = 3x + 7

Comparing this equation with y = mx + c

∴ Slope (m1) = 3

Now, we have to convert equation (2) in general form of y = mx + c

2y + px = 3

2y = 3 – px

2y = – px + 3

y = – px + 3/2

y = -p/2 x +3/2

∴ Slope m2 = -p/2

Also, given that, two lines are perpendicular

We know that,

If two lines are perpendicular then the slopes of two lines are – 1.

m1 × m2 = -1

3 × -p/2 = -1

-3p/2 = -1

P = 2/3

Here is your solution of Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14D

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Updated: November 25, 2021 — 2:47 pm